Exercise 7.1 Practice
Congruence of Triangles (SAS, ASA, AAS)
Q1: Quadrilateral Congruence
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In quadrilateral $ACBD$, $AC = AD$ and $AB$ bisects $\angle A$. Show that $\triangle ABC \cong \triangle ABD$. What can you say about $BC$ and $BD$?
In $\triangle ABC$ and $\triangle ABD$:
1. $AC = AD$ (Given)
2. $\angle CAB = \angle DAB$ (Since $AB$ bisects $\angle A$)
3. $AB = AB$ (Common side)
1. $AC = AD$ (Given)
2. $\angle CAB = \angle DAB$ (Since $AB$ bisects $\angle A$)
3. $AB = AB$ (Common side)
$\therefore \triangle ABC \cong \triangle ABD$ (By SAS Congruence Rule).
Conclusion about BC and BD:
Since the triangles are congruent, their corresponding parts are equal (CPCT).
$\therefore BC = BD$.
Since the triangles are congruent, their corresponding parts are equal (CPCT).
$\therefore BC = BD$.
Proved. BC = BD
Q2: ABCD Quadrilateral
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$ABCD$ is a quadrilateral in which $AD = BC$ and $\angle DAB = \angle CBA$. Prove that:
(i) $\triangle ABD \cong \triangle BAC$
(ii) $BD = AC$
(iii) $\angle ABD = \angle BAC$
(i) $\triangle ABD \cong \triangle BAC$
(ii) $BD = AC$
(iii) $\angle ABD = \angle BAC$
(i) In $\triangle ABD$ and $\triangle BAC$:
1. $AD = BC$ (Given)
2. $\angle DAB = \angle CBA$ (Given)
3. $AB = BA$ (Common)
$\therefore \triangle ABD \cong \triangle BAC$ (By SAS Rule).
1. $AD = BC$ (Given)
2. $\angle DAB = \angle CBA$ (Given)
3. $AB = BA$ (Common)
$\therefore \triangle ABD \cong \triangle BAC$ (By SAS Rule).
(ii) BD = AC:
Since $\triangle ABD \cong \triangle BAC$, by CPCT, $BD = AC$.
Since $\triangle ABD \cong \triangle BAC$, by CPCT, $BD = AC$.
(iii) $\angle ABD = \angle BAC$:
By CPCT, corresponding angles are equal.
$\therefore \angle ABD = \angle BAC$.
By CPCT, corresponding angles are equal.
$\therefore \angle ABD = \angle BAC$.
Proved all parts.
Q3: Perpendicular Bisector
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$AD$ and $BC$ are equal perpendiculars to a line segment $AB$. Show that $CD$ bisects $AB$.
Let $CD$ intersect $AB$ at $O$. We need to show $OA = OB$.
In $\triangle OAD$ and $\triangle OBC$:
1. $\angle OAD = \angle OBC = 90^\circ$ (Given perpendiculars)
2. $\angle AOD = \angle BOC$ (Vertically Opposite Angles)
3. $AD = BC$ (Given)
1. $\angle OAD = \angle OBC = 90^\circ$ (Given perpendiculars)
2. $\angle AOD = \angle BOC$ (Vertically Opposite Angles)
3. $AD = BC$ (Given)
$\therefore \triangle OAD \cong \triangle OBC$ (By AAS Rule).
By CPCT, $OA = OB$.
Thus, $CD$ bisects $AB$.
Thus, $CD$ bisects $AB$.
Proved.
Q4: Parallel Intersections
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$l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$. Show that $\triangle ABC \cong \triangle CDA$.
Since $l \parallel m$ and $p \parallel q$, quadrilateral $ABCD$ is a parallelogram.
Thus, $AB \parallel DC$ and $BC \parallel AD$.
Thus, $AB \parallel DC$ and $BC \parallel AD$.
In $\triangle ABC$ and $\triangle CDA$:
1. $\angle BAC = \angle DCA$ (Alternate Interior Angles, $AB \parallel DC$)
2. $\angle BCA = \angle DAC$ (Alternate Interior Angles, $BC \parallel AD$)
3. $AC = CA$ (Common side)
1. $\angle BAC = \angle DCA$ (Alternate Interior Angles, $AB \parallel DC$)
2. $\angle BCA = \angle DAC$ (Alternate Interior Angles, $BC \parallel AD$)
3. $AC = CA$ (Common side)
$\therefore \triangle ABC \cong \triangle CDA$ (By ASA Rule).
Proved.
Q5: Angle Bisector
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Line $l$ is the bisector of an angle $\angle A$ and $B$ is any point on $l$. $BP$ and $BQ$ are perpendiculars from $B$ to the arms of $\angle A$. Show that:
(i) $\triangle APB \cong \triangle AQB$
(ii) $BP = BQ$ or $B$ is equidistant from the arms of $\angle A$.
(i) $\triangle APB \cong \triangle AQB$
(ii) $BP = BQ$ or $B$ is equidistant from the arms of $\angle A$.
(i) In $\triangle APB$ and $\triangle AQB$:
1. $\angle APB = \angle AQB = 90^\circ$ (Given perpendiculars)
2. $\angle PAB = \angle QAB$ ($l$ bisects $\angle A$)
3. $AB = AB$ (Common side)
$\therefore \triangle APB \cong \triangle AQB$ (By AAS Rule).
1. $\angle APB = \angle AQB = 90^\circ$ (Given perpendiculars)
2. $\angle PAB = \angle QAB$ ($l$ bisects $\angle A$)
3. $AB = AB$ (Common side)
$\therefore \triangle APB \cong \triangle AQB$ (By AAS Rule).
(ii) BP = BQ:
By CPCT, $BP = BQ$.
This means $B$ is equidistant from the arms of $\angle A$.
By CPCT, $BP = BQ$.
This means $B$ is equidistant from the arms of $\angle A$.
Proved.
Q6: Complex Figure
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In the figure below, $AC = AE$, $AB = AD$ and $\angle BAD = \angle EAC$. Show that $BC = DE$.
Given $\angle BAD = \angle EAC$.
Add $\angle DAC$ to both sides:
$\angle BAD + \angle DAC = \angle EAC + \angle DAC$
$\Rightarrow \angle BAC = \angle DAE$.
Add $\angle DAC$ to both sides:
$\angle BAD + \angle DAC = \angle EAC + \angle DAC$
$\Rightarrow \angle BAC = \angle DAE$.
In $\triangle ABC$ and $\triangle ADE$:
1. $AB = AD$ (Given)
2. $\angle BAC = \angle DAE$ (Proved above)
3. $AC = AE$ (Given)
1. $AB = AD$ (Given)
2. $\angle BAC = \angle DAE$ (Proved above)
3. $AC = AE$ (Given)
$\therefore \triangle ABC \cong \triangle ADE$ (By SAS Rule).
By CPCT, $BC = DE$.
Proved.
Q7: Midpoint and Angles
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$AB$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $AB$ such that $\angle BAD = \angle ABE$ and $\angle EPA = \angle DPB$. Show that:
(i) $\triangle DAP \cong \triangle EBP$
(ii) $AD = BE$
(i) $\triangle DAP \cong \triangle EBP$
(ii) $AD = BE$
Given $\angle EPA = \angle DPB$.
Add $\angle EPD$ to both sides:
$\angle EPA + \angle EPD = \angle DPB + \angle EPD$
$\Rightarrow \angle APD = \angle BPE$.
Add $\angle EPD$ to both sides:
$\angle EPA + \angle EPD = \angle DPB + \angle EPD$
$\Rightarrow \angle APD = \angle BPE$.
In $\triangle DAP$ and $\triangle EBP$:
1. $\angle DAP = \angle EBP$ (Given $\angle BAD = \angle ABE$)
2. $AP = BP$ (P is midpoint of AB)
3. $\angle APD = \angle BPE$ (Proved above)
1. $\angle DAP = \angle EBP$ (Given $\angle BAD = \angle ABE$)
2. $AP = BP$ (P is midpoint of AB)
3. $\angle APD = \angle BPE$ (Proved above)
$\therefore \triangle DAP \cong \triangle EBP$ (By ASA Rule).
(ii) AD = BE:
By CPCT, $AD = BE$.
By CPCT, $AD = BE$.
Proved.
Q8: Right Triangle Properties
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In right triangle $ABC$, right angled at $C$, $M$ is the mid-point of hypotenuse $AB$. $C$ is joined to $M$ and produced to a point $D$ such that $DM = CM$. Point $D$ is joined to point $B$. Show that:
(i) $\triangle AMC \cong \triangle BMD$
(ii) $\angle DBC$ is a right angle
(iii) $\triangle DBC \cong \triangle ACB$
(iv) $CM = \frac{1}{2} AB$
(i) $\triangle AMC \cong \triangle BMD$
(ii) $\angle DBC$ is a right angle
(iii) $\triangle DBC \cong \triangle ACB$
(iv) $CM = \frac{1}{2} AB$
(i) $\triangle AMC \cong \triangle BMD$:
1. $AM = BM$ (M is midpoint)
2. $\angle AMC = \angle BMD$ (Vertically Opposite)
3. $CM = DM$ (Given)
$\therefore \triangle AMC \cong \triangle BMD$ (SAS Rule).
1. $AM = BM$ (M is midpoint)
2. $\angle AMC = \angle BMD$ (Vertically Opposite)
3. $CM = DM$ (Given)
$\therefore \triangle AMC \cong \triangle BMD$ (SAS Rule).
(ii) $\angle DBC = 90^\circ$:
From (i), $\angle MAC = \angle MBD$ (CPCT). These are alternate interior angles, so $AC \parallel DB$.
Since $AC \parallel DB$ and $BC$ is transversal, $\angle ACB + \angle DBC = 180^\circ$ (Consecutive Interior).
Given $\angle ACB = 90^\circ$, so $\angle DBC = 90^\circ$.
From (i), $\angle MAC = \angle MBD$ (CPCT). These are alternate interior angles, so $AC \parallel DB$.
Since $AC \parallel DB$ and $BC$ is transversal, $\angle ACB + \angle DBC = 180^\circ$ (Consecutive Interior).
Given $\angle ACB = 90^\circ$, so $\angle DBC = 90^\circ$.
(iii) $\triangle DBC \cong \triangle ACB$:
1. $DB = AC$ (CPCT from i)
2. $\angle DBC = \angle ACB = 90^\circ$
3. $BC = CB$ (Common)
$\therefore \triangle DBC \cong \triangle ACB$ (SAS Rule).
1. $DB = AC$ (CPCT from i)
2. $\angle DBC = \angle ACB = 90^\circ$
3. $BC = CB$ (Common)
$\therefore \triangle DBC \cong \triangle ACB$ (SAS Rule).
(iv) $CM = \frac{1}{2} AB$:
From (iii), $DC = AB$ (CPCT).
Since $M$ is midpoint of $DC$, $CM = \frac{1}{2} DC$.
$\therefore CM = \frac{1}{2} AB$.
From (iii), $DC = AB$ (CPCT).
Since $M$ is midpoint of $DC$, $CM = \frac{1}{2} DC$.
$\therefore CM = \frac{1}{2} AB$.
Proved all parts.