D IS A POINT ON THE SIDE BC OF A TRIANGLE ABC SUCH THAT /ADC=/BAC.SHOW THAT CA2 CB.CD.
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IN ΔABC AND ΔADC
∠C = ∠C (COMMON)
∠BAC =∠ADC
∴ ΔABC ≈ ΔADC (AA CRITERIA)
∴AC/DC = BC/AC (CPST)
∴CA²=CB*CD
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∠C = ∠C (COMMON)
∠BAC =∠ADC
∴ ΔABC ≈ ΔADC (AA CRITERIA)
∴AC/DC = BC/AC (CPST)
∴CA²=CB*CD
HOPE IT WILL HELP YOU
MARK ME BRAINIEST
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