Question

in figure triangle ABC is a right angled at a q and R are points on line BC and P is a point such that QP parallel to AC and RP parallel to ab find angle p

Answer 1

Angle P would be 90°

Step-by-step explanation:

In the given diagram,

ABC is a a right angled at A,

Q and R are points on line BC,

There is also a point P such that QP║AC and RP ║ AB,

∵ QP║AC

By Alternative interior angle theorem,

m∠ACB = m∠PQR,

Similarly,  RP ║ AB ⇒ m∠ABC = m∠PRQ

By AA similarity postulate,

$\triangle ABC\sim \triangle PRQ$

∵ In two similar triangles corresponding angles are equal in measure,

⇒ m∠CAB = m∠RPQ

Hence, Angle P would be 90°

#Learn more:

In a triangle ABC , E and F are the mid points of AC and AB respectively . The altitude AP to BC intersect EF at Q .Provet that AQ = QP.

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