84 is 12% of what number
2 answers:
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Given: An Isosceles trapezoid EFGH in which EF =GH
To prove: ΔFHE ≅ ΔGEH
Proof: In Isosceles trapezoid EFGH, Considering two triangles ΔFHE and ΔGEH
1. FE ≅ G H → [ Given]
2. ∠H = ∠E
→ Draw GM⊥HE and FN ⊥EH, and In Δ GMH and ΔFNE,
GH=FE [Given]
∠M+∠N=180° so GM║FN and GF║EH, So GFMN is a rectangle.]
∴ GM =FN [opposite sides of rectangle]
∠GMH = ∠FNE [ Each being 90°]
Δ GMH ≅ ΔFNE [ Right hand side congruency]
→∠H =∠E [CPCT]
→ Side EH is common i.e EH ≅ EH .
→ΔFHE ≅ ΔGEH. [SAS]
