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heena

· started a discussion

· 1 Months ago

solution please

In trianlgle abc, angle c is obtuse.the bisectors of exterior angles at A and B meet BC and AC produced at D and E respectively. If AB=AD=BE then angle ACB is...??????

please explain with figure

heena

· commented

· 1 Months ago

thanks

Knowledge Expert

· commented

· 1 Months ago

From the question,
AB = AD = BE
Let, D = x and E = y

In triangle ABD,
ADB = ABD = x ( since AB = AD)

Similarly, in triangle ABE,
AEB = EAB = y

In triangle ABD,
PAD = ADB + ABD = 2x (exterior angle of an triangle)

Since, AD is bisector of A, PAD = CAD = 2x
Similarly, BE is bisector of B. So,
QBE = CBE = 2y

In triangle ABD,
ADB + ABD +BAD = 180
⇒ x + x + CAD + CAB = 180
⇒ 2x + 2x + y = 180
⇒ 4x + y = 180 -----eq 1

Similarly, for triangle ABE,
4y + x = 180 -------eq 2

Solving the two equations simultaneously
x = 36 and y = 36

Hence, in triangle ABC,
x + y + ACB = 180
⇒36 + 36 + ACB = 180
⇒ACB = 108
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