Question

please answer this question if you know the answer then only answer else reported​

Answer 1

Step-by-step explanation:

★Given -

• $∠1 = ∠2$
• $\angle 3 = \angle 4$

• $\rm{ \angle 1 = \angle 2 = \dfrac{1}{2} \angle ABC}$

• $\rm{ \angle 3= \angle 4 = \dfrac{1}{2} \angle ACD}$

★To prove -

• $\rm{ \angle BTC = \dfrac{1}{2} \angle BAC}$

★Solution -

In ∆ABC,

${\longmapsto\rm{ \bf∠A + ∠B = ∠ACD \: \: \: \: \: \: \: ( exterior \: angle \: property)}}$

Dividing both sides by 1/2,

${\longmapsto\rm{ \bf \dfrac{1}{2} ∠A + \dfrac{1}{2} ∠B = \dfrac{1}{2} ∠ACD }}$

${\longmapsto\rm{ \bf \dfrac{1}{2} ∠A + \dfrac{1}{2} ∠B = ∠4 \: \: \: \: \: (1) }}$

Now, in ∆TBC,

${\longmapsto\rm{ \bf∠2 + ∠T = ∠4 \: \: \: \: \: \: \: ( exterior \: angle \: property)}}$

${\longmapsto\rm{ \bf \dfrac{1}{2} \angle B + ∠T = ∠4 \: \: \: \: \: \: \: (2) }}$

Equating (1) and (2),

${\longmapsto\rm{ \bf \dfrac{1}{2} ∠A + \cancel{\dfrac{1}{2} ∠B} = \cancel{\dfrac{1}{2} \angle B} + ∠T }}$

${\longmapsto\rm{ \bf \dfrac{1}{2} ∠A = ∠T }}$

$\longmapsto \rm{ \bf \dfrac{1}{2} \angle \: BAC = \angle BTC }$

or

${\longmapsto\rm{ \bf \orange{ \angle BTC = \dfrac{1}{2} \angle BAC \: \green{ , hence \: proved}}}}$