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We know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to each other and to the whole triangle.
In ΔPQR we have,
∠QPR = 90° and PM ⊥ QR
In ΔPQR and ΔMQP
∠QPR = ∠QMP = 90°
∠PQR = ∠MQP (commom angle)
⇒ ΔPQR ~ ΔMQP (AA Similarity) --------(1)
In ΔPQR and ΔMPR
∠QPR = ∠PMR = 90°
∠PRQ = ∠PRM (commom angle)
⇒ ΔPQR ~ ΔMPR (AA Similarity) --------(2)
From equation (1) and (2)
ΔMQP ~ ΔMPR
PM / MR = QM / PM (corresponding sides of similar triangles are proportional)
Answers & Comments
Answer:
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Verified answer
Explanation:
[tex]\underline{\underline{\pink{\huge\sf ✨ Solution✨}}}[/tex]
We know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to each other and to the whole triangle.
In ΔPQR we have,
∠QPR = 90° and PM ⊥ QR
In ΔPQR and ΔMQP
∠QPR = ∠QMP = 90°
∠PQR = ∠MQP (commom angle)
⇒ ΔPQR ~ ΔMQP (AA Similarity) --------(1)
In ΔPQR and ΔMPR
∠QPR = ∠PMR = 90°
∠PRQ = ∠PRM (commom angle)
⇒ ΔPQR ~ ΔMPR (AA Similarity) --------(2)
From equation (1) and (2)
ΔMQP ~ ΔMPR
PM / MR = QM / PM (corresponding sides of similar triangles are proportional)
⇒ PM2 = QM.MR
Hence proved.