Answer with explanation:
In ΔABC, Right Angled at B.
Draw , B D ⊥ AC
Also,1. ∠B DC≅∠A DB
2.∠B CA≅∠DCB
3.∠B AC≅∠BAD
4.∠D BC≅∠B AC
To Prove: →AB²+BC²=AC²
Solution:
In ΔABC and ΔB DC
B D ⊥ AC→→By Construction
∠ABC=∠BDC→→[Each being 90°]
∠ACB=∠DCB→→[Reflexive angles , →Reflexive Property of Congruence]
→→→ΔABC~ΔB DC→ [AA criterion for similarity]
When, triangles are similar,corresponding angles are equal and corresponding sides are proportional.
[tex]\frac{BC}{AC}=\frac{DC}{BC}[/tex]
BC²=AC×DC -------------------(1)
In ΔADB and ΔB AC
B D ⊥ AC→→By Construction
∠ABC=∠ADB→→[Each being 90°]
∠DAB=∠CAB→→[Reflexive angles , →Reflexive Property of Congruence]
→→→ΔADB~ΔB AC→ [AA criterion for similarity]
When, triangles are similar,corresponding angles are equal and corresponding sides are proportional.
[tex]\frac{BA}{AC}=\frac{DA}{BA}[/tex]
AB²=AC × AD -------------------(2)
Adding (1) and (2), we get
AB²+BC²=AC × AD +AC × CD
AB²+BC²=AC ×(AD +CD)
AB²+BC²=AC × AC
AB²+BC²=AC²
Hence proved.
