In Delta PQR,X is the midpoint of QR. XY and | vedclass

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

Question

(N/A) Given: In $\Delta PQR$,$X$ is the midpoint of $QR$,so $QX = XR$. $XY \perp PQ$ and $XZ \perp PR$. Also,$XY = XZ$.Step $1$: Consider $\Delta QXY$

Vedclass · Accepted Answer

Vedclass · Answer

(N/A) Given: In $\Delta PQR$,$X$ is the midpoint of $QR$,so $QX = XR$. $XY \perp PQ$ and $XZ \perp PR$. Also,$XY = XZ$.
Step $1$: Consider $\Delta QXY$ and $\Delta RXZ$.
Step $2$: In these two triangles:
$1$. $\angle XYQ = \angle XZR = 90^{\circ}$ (Given as altitudes).
$2$. $XY = XZ$ (Given).
$3$. $QX = XR$ (Given as $X$ is the midpoint of $QR$).
Step $3$: By $RHS$ congruence criterion,$\Delta QXY \cong \Delta RXZ$.
Step $4$: By $CPCT$,$\angle Q = \angle R$.
Step $5$: In $\Delta PQR$,since $\angle Q = \angle R$,the sides opposite to these angles must be equal,i.e.,$PR = PQ$.
Conclusion: Since two sides of $\Delta PQR$ are equal,$\Delta PQR$ is an isosceles triangle.

In $\Delta PQR$,$X$ is the midpoint of $QR$. $XY$ and $XZ$ are altitudes from $X$ on $PQ$ and $PR$ respectively. If $XY = XZ$,then prove that $\Delta PQR$ is an isosceles triangle.

Similar Questions

State whether each of the following statements is true or false:
$(1)$ In $\Delta ABC$ and $\Delta PQR$,$AB = PQ$,$BC = PR$ and $\angle B = \angle P$,then $\Delta ABC \cong \Delta PQR$.
$(2)$ The exterior angles at two vertices of any triangle cannot both be acute.

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

If the bisector of an angle of a triangle also bisects the opposite side,prove that the triangle is isosceles.

$ABC$ is an isosceles triangle with $AB = AC$. $BD$ and $CE$ are its two medians. Show that $BD = CE$.

In $\Delta ABC$, $\angle B = \angle C$, $AB = 5 \text{ cm}$, and $BC = 8 \text{ cm}$. Then, the perimeter of $\Delta ABC$ is: (in $\text{ cm}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module