In ` triangle ` ABC ,
` angle ACB = 90^(@)` ,
` CD bot AB` and
` DE bot CB`
Prove that ` CD^(2) xxAC = AD xx AB xxDE ` .

In DeltaABC,angleACB=90^(@) seg CD bot seg AB seg DE bot seg CB. Show that: CD^(2)xxAC=ADxxABxxDE

AB botBC and DE bot AC . Prove that DeltaABC ~ DeltaAED .

In the given figure, angle ACB=90^(@) and CD bot AB . Prove that CD^(2)=BD*AD

In Delta ABC, angle B = 90^(@) and BD bot AC. If AC = 9 cm and AD = 3 cm then BD is equal to

In a right triangle ABC, in which angleC = 90^@ amd CD bot AB. If BC =a, CA=b, AB=c and CD=p, then

In triangle ABC , angle A = 90^@ , AD is the bisector of angle A meeting BC at D and DE bot AC at E. If AB =10cm and AC = 15 cm, then the length of DE, in cm is: triangle ABC में angle A=90^@ , AD, angle A का द्विभाजक है, जो BC से D पर मिलता है, और DE bot AC से E पर मिलता है। यदि AB = 10 cm और AC= 15 cm है, तो DE की लंबाई(cm) है,

In Delta ABC, angle B = 90^(@) and BD bot AC . If AC = 9 cm and AD = 3 cm, then BD is equal to :

In Delta ABC, angle ACB = 90^(@) ," seg "CD bot side AB and seg CE is angle bisector of angle ACB Prove : (AD)/(BD) = (AE^(2))/(BE^(2))

In triangle ABC , angle A=90^@ . M is the midpoint of BC and D is a point on BC such that AD bot BC . If AB= 7 cm and AC = 24 cm, then AD:AM is equal to: triangle ABC में, angle A=90^@ है, M, BC का मध्य बिंदु है तथा D, BC पर स्थित ऐसा बिंदु है कि AD bot BC है | यदि AB= 7 सेमी, AC = 24 सेमी है, तो AD:AM किसके बराबर है ?