Given: BD bisects <B, BD is perpendicular to AC Prove: <A is congruent to <C

Question

Please help me! I am very confused with geometry!

Answer 1

I'll assume that ABC is a triangle, ΔABC. And BD is an altitude of ΔABC.

 

If BD bisects ∠B,  then ∠ABD ≅ ∠CBD, by definition of bisect. 

 

If BD is perpendicular to AC, then both angles ∠ADB and ∠CDB are right angles by definition of perpendicular. (Angle)

And ∠ADB ≅ ∠CDB because all right angles are congruent. (Angle)

And bd ≅ bd by reflexive property (Included side - between the two congruent angles)

 

             NOTE: Now you have angle-side-angle (ASA)

 

Therefore ΔABD ≅ ΔCBD by ASA

And ∠A ≅ ∠C by CPCTC (congruent parts of congruent triangles are congruent)

 

Think of proofs as a puzzle, and look at each angle and each side as it's own part. Then list, actually right down or mark, what you know about each part and identify any traits any two parts might share. Then pull together similarities to find out of you have AAS, ASA, SSS or SAS. And if not, what do you need to get to one of these and see if you can find additional similarities. 

 

I hope this helps!