Question

Write the complete proof in your paper homework and for online (only) complete the probing statement (if any) that is a part of your proof or related to it.

Answer 1

proof below

Explanation:

∠A ≅ ∠B

as ACB is a triangle

if ∠A ≅ ∠B then AC = BC

and CD ⊥ AB

so ∠CDA = ∠CDB = 90

now in ΔADC ---> ∠A + ∠CDA + ∠ACD = 180

lets assume ∠A = x = ∠B

x + 90 + ∠ACD = 180

∠ACD = 180 - 90 - x

∠ACD = 90 - x

now in ΔCDB ---> ∠B + ∠CDB + ∠BCD = 180

x + 90 + ∠BCD = 180

∠BCD = 180 - 90 - x

∠BCD = 90 - x

As ∠ACD = ∠BCD

CD bisects ∠ACB

Answer 2

Indirectly to proof:-

• <ACD=<BCD

Lets see

In ∆ADC and ∆BDC

As CD bisects AB

• AD=BD
• CD is common side
• <A=<B(Given)

Hence by SSA congruency the triangles are equal

<ACD=<BCD( proved)