In the adjoining figure, $A B = A C$ . If $P M ⊥ A B$ and $P N ⊥ A C$ , show that $P M \times P C = P N \times P B$ .

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Solution

Given $A B = A C$
$∠ A B C = ∠ A C B$ (Isosceles triangle property) (I)
In $△ P M B$ and $△ P N C$
$∠ M B P = ∠ N C P$ (From I)
$∠ P M B = ∠ P N C$ (each $90^{\circ}$ )
$∠ M P B = ∠ N P C$ (third angle)
Thus, $△ P M B \sim △ P N C$ (AAA rule)
hence, $\frac{P M}{P N} = \frac{P B}{P C}$ (Corresponding sides)
$P M \times P C = P B \times P N$

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Similar questions

State true or false:

△ A B C

∠ A

is obtuse and

A B = A C

P

is any point in side

B C

P M ⊥ A B

and

P N ⊥ A C .

Then,

P M \times P C = P N \times P B

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△ A P M ≅ △ A P N

P M = P N, P M ⊥ A B

and

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A point P is in the interior of angle BAC, such that P lies on the bisector of angle BAC. What can be said about the distance of PM if PN=2cm where PM and PN are perpendiculars from P on the lines BA and AC?

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In the adjoining figure, AB=AC. If PM⊥AB and PN⊥AC, show that PM×PC=PN×PB.

Given AB=AC∠ABC=∠ACB (Isosceles triangle property) (I)In △PMB and △PNC∠MBP=∠NCP (From I)∠PMB=∠PNC (each 90∘)∠MPB=∠NPC (third angle)Thus, △PMB∼△PNC (AAA rule)hence, PMPN=PBPC (Corresponding sides)PM×PC=PB×PN

In the adjoining figure, $A B = A C$ . If $P M ⊥ A B$ and $P N ⊥ A C$ , show that $P M \times P C = P N \times P B$ .