We have,
ABC is a right angle triangle.
In which ∠ACB=90o
And CD⊥ABandDE⊥AB
Where D is any point in AB and E is any point in CB.
Prove that:- CD2×AC=AD×AB×DE
Proof:-
InΔACBandΔADC
∠CAB≅∠DAC(Reflexive)
∠ACB≅∠ADC(Rightangle)
Thus, By AA similarity.
Then,
ΔACB∼ΔADC......(1)
⇒ACAD=ABAC
⇒AC2=AB×AD......(2)
Now, Similarly
ΔACB∼ΔCDB......(3)
Now,
InΔCEDandΔCDB
∠ECD≅∠DCB(Reflexive)
∠CED≅∠CDB(Rightangle)
By AA similarity.
ΔCED∼ΔCDB.......(4)
From equation (3) and (4) to,
ΔACB∼ΔCED......(5)
By equation (1) and (5) to, we get,
ΔADC∼ΔCED
ACCD=DCED
CD2=AC×DE......(6)
From equation (2) to,
AC2=AB×AD
On multiplying both side by CD2 and we get,
CD2×AC2=CD2×AB×AD
CD2×AC2=AC×DE×AB×AD
CD2×AC=DE×AB×AD
CD2×AC=AD×AB×DE
Hence
proved.