Line segments AB and CD bisect each other at | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(N/A) Given: $AB$ and $CD$ bisect each other at $P$. This implies $PA = PB$ and $PC = PD$.However,the problem states $PA = PD$ and $PB = PC$.Consider

Vedclass · Accepted Answer

Vedclass · Answer

(N/A) Given: $AB$ and $CD$ bisect each other at $P$. This implies $PA = PB$ and $PC = PD$.
However,the problem states $PA = PD$ and $PB = PC$.
Consider $\triangle APC$ and $\triangle BPD$:
$1$. $PA = PD$ (Given)
$2$. $\angle APC = \angle BPD$ (Vertically opposite angles)
$3$. $PC = PB$ (Given)
By $SAS$ (Side-Angle-Side) congruence criterion,$\triangle APC \cong \triangle BPD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$AC = BD$.

Line segments $AB$ and $CD$ bisect each other at $P$. If $PA = PD$ and $PB = PC$,prove that $AC = BD$.

Similar Questions

In $\Delta ABC$,$AB = AC$ and $\angle B = 75^{\circ}$,then $\angle C = \dots$ (in $^{\circ}$)

$ABC$ is a right triangle such that $AB = AC$ and the bisector of angle $C$ intersects the side $AB$ at $D$. Prove that $AC + AD = BC$.

$D$ is a point on the side $BC$ of a $\triangle ABC$ such that $AD$ bisects $\angle BAC$. Then

In $\Delta ABC$,$AB = AC$. If $\angle A = 50^{\circ}$,then find $\angle B$ and $\angle C$.

There can be $\ldots \ldots \ldots$ one-one correspondences between two triangles.

Vedclass Test Series

Exam Paper Generator

Online Exam Module