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nikdorinn [45]
4 years ago
8

What is the missing statement in the proof? Scroll down to see the entire proof. ∠BDC≅∠ADB ∠BCA≅∠DCB ∠BAC≅∠BAD ∠DBC≅∠BAC Done Gi

ven: ΔABC with m∠ABC = 90° (view diagram) Prove: AB2 + BC2 = AC2 Statement Reason 1. Draw BD¯¯¯¯¯⊥AC¯¯¯¯¯ (view diagram). construction 2. ∠ABC≅∠BDC Angles with the same measure are congruent. 3. Reflexive Property of Congruence 4. ΔABC~ΔBDC AA criterion for similarity 5. BCDC=ACBC Corresponding sides of similar triangles are proportional. 6. BC2 = AC × DC cross multiplication 7. ∠ABC≅∠ADB Angles with the same measure are congruent. 8. ∠BAC≅∠DAB Reflexive Property of Congruence 9. ΔABC~ΔADB AA criterion for similarity 10.ABAD=ACAB Corresponding sides of similar triangles are proportional. 11. AB2 = AC × AD cross multiplication 12. AB2 + BC2 = AC × AD + AC × DC addition 13. AB2 + BC2 = AC(AD + DC)

Mathematics
1 answer:
denpristay [2]4 years ago
7 0

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.

\frac{BC}{AC}=\frac{DC}{BC}

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.

\frac{BA}{AC}=\frac{DA}{BA}

 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.

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