In a triangle ABC, AB=AC. D is any point on BC. Show that AB^{ 2 }-AD ^{ } ^{ 2 } ^{ } = ^{ BD.CD }

1. Show that the area of a rhombus on hypotenuse of a right angled triangle, with one of the angles as 60 is equal to the sum of areas of rhombuses with one of their angles as 60 drawn on others sides.

2. BO and CO bisect angle B and angle C of triangle ABC. AO produced meets BC at P. Show:-

· AB*OP=BP*AO

· AC*OP=CP*AO

· AB*PC=AC*BP

· AP bisects angle BAC.

In a triangle ABC, AB=AC. D is any point on BC. Show that AB^{ 2 }-AD ^{2 }= BD . CD

Solution:

To prove: AB^{ 2 }-AD ^{2 }= BD . CD

Construction: Draw AE perpendicular to BC.

Proof: In tr. AEB and tr. AEC,

AB= AC [given]

∠AEB = ∠AEC = 90°

AE= AE [common]

∴ΔAEB ΔAEC [By RHS]

⇒BE= EC[cpct]

in tr. ABE,

∠E= 90°

on subtracting (ii) from (i). we get,

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