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In Delta ABC, AB = AC. A circle passing ...

In `Delta ABC, AB = AC`. A circle passing through B touches side AC at its midpoint P and intersects side AB at Q then prove BQ = 3AQ
Construction `:` Draw seg BP and seg PQ

Text Solution

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In `Delta APQ ` and `Delta ABP`,
`/_ APQ ~= /_ABP` …(Tangent secant theorem )
`/_ PAQ ~= /_ BAP` …(Common angle )
`:. Delta APQ ~ Delta ABP ` …(AA test of similarity )
`:. (AP )/(AB) = ( AQ)/(AP ) ` ….(c.s.s.t.)
`:. ((1)/(2) AC )/( AB)= (AQ)/((1)/(2)AC)= ` ...(` :'` P is the midpoint of seg AC )
`:. ((1)/(2) AB)/(AB) = (AQ)/((1)/(2)AB) ` ...(` :. AB = AC )`
`:. AQ = (1)/(2) xx (1)/(2) AB `
`:. AQ = (1)/(4) AB `
`:. AB = 4AQ `
`BQ+AQ = AB ` ...(A-Q-B)`
`BQ + AQ = 4AQ `
`:. BQ = 4AQ - AQ `
`:. BQ = 3AQ `
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