Question

2ABC is a triangle. The bisector of the angle BCA
meets AB in X. A point Y lies on CX such that
AX= AY. Prove that ZCAY = ZABC. (ICSE)​

Answer 1

$\huge\underline\bold\red{QUESTION ???}$

ABC is a triangle. The bisector of the angle BCA meets AB in X. A point Y lies on CX such that AX= AY. Prove that ∠CAY = ∠ABC.

$\sf\underline{\underline{\</strong><strong>b</strong><strong>l</strong><strong>u</strong><strong>e</strong><strong>{</strong><strong>G</strong><strong>I</strong><strong>V</strong><strong>E</strong><strong>N</strong><strong>:</strong><strong>-</strong><strong> \: }}}$

• ABC is a triangle.
• The bisector of the angle BCA meets AB in X.
• A point Y lies on CX such that AX= AY.

$\sf\underline{\underline{\green{TO \: PROVE:- \: }}}$

∠CAY = ∠ABC

$\huge\underline\mathbb{\red S\pink {0} \purple {L} \blue {UT} \orange {1}\green {ON :}}$

$\sf\underline{\underline{\red{PROOF:- \: }}}$

In ∆ ABC,

CX is the angle bisector of ∠C

⇒ ∠ACY = ∠BCX ....... (1)

and AX = AY

So,

In ∆ AXY

∠AXY = ∠AYX (Angles opposite to equal sides) .......... (2)

Now,

∠XYC = ∠AXB = 180° (straight line)

⇒∠AYX + ∠AYC = ∠AXY + ∠BXY

⇒∠AYC + ∠BXY (from (2)) ........ (3)

Also In ∆ AYC and ∆ BXC

∠AYC + ∠YCA + ∠CAY = ∠BXC + ∠BCX + ∠XBC = 180° (Angle sum property)

⇒ ∠CAY = ∠XBC ......(from (1) and (3))

⇒ ∠CAY = ∠ABC

$\sf\underline{\underline{\green{<strong>HENCE</strong> \: <strong>PROVED</strong> \: }}}$

Answer 2

Given :

• ABC is a triangle
• The bisector of the angle BCA meets AB in X
• A point Y lies on CX such that AX = AY

To prove :

• ∠ CAY = ∠ ABC

Proof :

Since, AX = AY

→ ∴ ∠ AXY = ∠ AYX   [ angles opposite to equal sides ] ____equation (1)

Now, since AXB and XYC are lines, therefore, by linear pair

→ ∠ BXY + ∠ AXY = ∠ AYX + ∠ AYC = 180°

By equation (1)

→ ∠ BXY + ∠ AXY = ∠ AXY + ∠ AYC

→ ∠ BXY = ∠ AYC   _____equation (2)

Now,

∵ CX is bisector of angle ACB

∴ ∠ BCX = ACY   _____equation (3)

Consider Δ BCX and Δ ACY,

by angle sum property of triangle

→ ∠ XBC + ∠ BCX + ∠ CXB = ∠ CAY + ∠ ACY + ∠ AYC = 180°

By equation (1) and (2)

→ ∠ XBC + ∠ BCX + ∠ BXY = ∠ CAY + ∠ BCX + ∠ BXY

→ ∠ XBC = ∠ CAY

therefore,

→ ∠ ABC = ∠ CAY

PROVED .

See the attachment for figure.