Can anyone solve question #15
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Answer:
Step-by-step explanation:
Given: ∠ACB=90° and CD⊥AB.
To prove:
Solution:
In ΔACD and ΔABC, we have
∠ADC=∠ACB=90°
∠CAD=∠BAC(Common)
Thus, by AA similarity, ΔACD is similar to ΔABC.
Hence,
⇒ (1)
Similarly, ΔDCB is similar to ΔCAB.
⇒
⇒ (2)
Now, dividing (2) by (1), we get
⇒
Hence proved.