Question

In any ∆ABC, E and D are interior points of AC and BC
respectively, AF bisects 6 CAD and BF bisects 6 CBE.
Prove that m AEB+ m ADB = 2m AF B.
[angles not areas]

Answer 1

Solution :-

In ∆AFH we have,

→ ∠AFH + ∠AHF + ∠FAH = 180° { By angle sum property.}

or,

→ ∠AFB + ∠AHF + ∠FAD = 180° -------- Eqn.(1)

Now, in ∆BFG we have,

→ ∠BFG + ∠BGF + ∠GBF = 180° { By angle sum property.}

or,

→ ∠AFB + ∠BGF + ∠FBE = 180° -------- Eqn.(2)

Now, in ∆AEG we have,

→ ∠AEG + ∠AGE + ∠EAG = 180° { By angle sum property.}

also,

→ ∠EAG = ∠CAF = ∠FAD { since AF bisects ∠A .}

and,

→ ∠AGE = ∠BGF { Vertically opposite angles.}

and,

→ ∠AEG = ∠AEB

So,

→ ∠AEB + ∠BGF + ∠FAD = 180° -------- Eqn.(3)

Similarly, in ∆BDH we have,

→ ∠BDH + ∠BHD + ∠HBD = 180° { By angle sum property.}

also,

→ ∠CBF = ∠FBE { since BF bisects ∠B .}

and,

→ ∠BHD = ∠FHA { Vertically opposite angles.}

and,

→ ∠BDH = ∠ADB

So,

→ ∠ADB + ∠FBE + ∠FHA = 180° -------- Eqn.(4)

Now, Subtracting sum of Eqn.(3) and Eqn.(4) from the sum of Eqn.(1) and Eqn.(2) we get,

→ [∠AFB + ∠AHF + ∠FAD + ∠AFB + ∠BGF + ∠FBE] - [∠AEB + ∠BGF + ∠FAD + ∠ADB + ∠FBE + ∠FHA] = (180° + 180°) - (180° + 180°)

→ 2•∠AFB - ∠AEB - ∠ADB = 0

→ ∠AEB + ∠ADB = 2•∠AFB (Proved.)

Learn more :-

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