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$ \triangle \mathrm{ABC} $ and $ \triangle \mathrm{DBC} $ are two isosceles triangles on the same base $ B C $ and vertices $ A $ and $ D $ are on the same side of $ \mathrm{BC} $ (see Fig. 7.39). If $ \mathrm{AD} $ is extended to intersect $ \mathrm{BC} $ at $ \mathrm{P} $, show that
(i) $ \triangle \mathrm{ABD} \equiv \triangle \mathrm{ACD} $
(ii) $ \triangle \mathrm{ABP} \cong \triangle \mathrm{ACP} $
(iii) $ \mathrm{AP} $ bisects $ \angle \mathrm{A} $ as well as $ \angle \mathrm{D} $.
(iv) AP is the perpendicular bisector of $ \mathrm{BC} $.
"
Given:
$\triangle ABC$ and $\triangle DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$. If $AD$ is extended to intersect $BC$ at $P$.
To do:
We have to show:
(i) $\triangle ABD \cong \triangle ACD$
(ii) $\triangle ABP \cong \triangle ACP$(iii) $AP$ bisects $\angle A$ as well as $\angle D$.
(iv) $AP$ is the perpendicular bisector of $BC$.
Solution:
(i) We know that,
The side-Side-Side congruence rule states that if three sides of one triangle are equal to three corresponding sides of another triangle, then the triangles are congruent.
Let us consider $\triangle ABD$ and $\triangle ACD$
Given,
$\triangle ABC$ and $\triangle DBC$ are isosceles triangles,
This implies,
$AB=AC$ and $BD=CD$
Since $AD$ is the common side
$AD=AD$
Therefore,
$\triangle ABD \cong \triangle ACD$
(ii) Let us consider $\triangle ABP$ and $\triangle ACP$
Given,
$\triangle ABC$ is isosceles,
This implies,
$AB=AC$
Since $AP$ is the common side
We get,
$AP=AP$
We also know
From corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding angles and sides must be equal.
Therefore,
$\angle PAB=\angle PAC$.
Therefore,
According to the Rule of Side-Angle-Side Congruence:
Triangles are said to be congruent if any pair of corresponding sides and their included angles are equal in both triangles.
Hence, $\triangle ABP \cong \triangle ACP$.
(iii) We know that,
From corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding angles and sides must be equal.
Therefore,
$\angle PAB=\angle PAC$ (since $\triangle ABD \cong \triangle ACD$)
Given,
$AP$ bisects $\angle A$...(i)
Let us consider $\triangle BPD$ and $\triangle CPD$
We also know that,
The side-Side-Side congruence rule states that if three sides of one triangle are equal to three corresponding sides of another triangle, then the triangles are congruent.
Since $PD$ is the common side.
We get, $PD=PD$
As $\triangle DBC$ is isosceles We get,
$BD=CD$
and by CPCT we know that as $\triangle ABP \cong \triangle ACP$
Therefore we get,
$\triangle BPD \cong \triangle CPD$
Hence, $\angle BDP=\angle CDP$ by CPCT....(ii)
Now, by comparing (i) and (ii) We can say that $AP$ bisects $\angle A$ as well as $\angle D$.
(iv) Let us consider $\triangle BPD$ and $\triangle CPD$
We know that,
From corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding angles and sides must be equal.
Therefore,
$\angle BPD=\angle CPD$
and $BP=CP$ ....(i)
We also know that,
The sum of the angles of a straight line is $180^o$
$\angle BPD+\angle CPD=180^o$
Since $\angle BPD=\angle CPD$
We get,
$2\angle BPD=180^o$
$\angle BPD=\frac{180^o}{2}$
$\angle BPD=90^o$.....(ii)
From (i) and (ii) we can say that,
$AP$ is the perpendicular bisector of $BC$.