Question

PQR is a triangle whose PQ=PR, S is the mid point of PQ and right angle bisector of PQ intersects extended base QR at point T. Prove that PQ^2=QR*QT

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The given question is PQR is a triangle whose PQ=PR, S is the midpoint of PQ and the right angle bisector of PQ intersects extended base QR at point T.

we have to prove that

${pq}^{2} = qr \times qt$

The data given in the question is

PQR is an isosceles triangle where PQ=PR.

S is the midpoint of PQ.

ST is perpendicular to PQ.

A line segment along a midpoint is drawn as ST.

join PT and RT.

From the triangle PST and triangle QST we have

PS will be equal to SQ (by midpoint theorem).

the angle of TSP will be equal to the angle of TSQ where ST is perpendicular to PQ.

ST is a common side.

The triangle PST is equivalent to the triangle QST by the side angle side theorem.

PT = QT (by corresponding sides of the congruent triangle are equal.)

angle TPQ and angle PQT are equal ( by corresponding angles of the congruent triangles are equal.

From ∆ PQR and ∆ TPQ and we get

angle TPQ and angle PQT is equal

angle PQR and angle PQT is equal by the concept of a common angle.

∆PQR ~∆ TPQ. this is the similarity to the triangle.

$\frac{pq}{tp} = \frac{qr}{pq} = \frac{pq}{tq}$

The above form can also be determined as

$pq \ \times pq \: = \: qr \: \times tq$

Therefore, the concept is proved.

# spj2

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