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Question
The following figure shows a triangle ABC in which AD and BE are perpendiculars to BC and AC respectively.
Show that:
- ΔADC ∼ ΔBEC
- CA × CE = CB × CD
- ΔABC ~ ΔDEC
- CD × AB = CA × DE
Solution
i. ∠ADC = ∠BEC = 90°
∠ACD = ∠BCE ...(Common)
ΔADC ∼ ΔBEC ...(AA similarity)
ii From part (i),
`(AC)/(BC) = (CD)/(EC)` ...(1)
`=>` CA × CE = CB × CD
iii. In ΔABC and ΔDEC,
From (1),
`(AC)/(BC) = (CD)/(EC) => (AC)/(CD) = (BC)/(EC)`
∠DCE = ∠BCA ...(Common)
ΔABC ~ ΔDEC ...(SAS similarity)
iv. From part (iii),
`(AC)/(DC) = (AB)/(DE)`
`=>` CD × AB = CA × DE
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