In AABC, D and E are two points on the sides AB and AC respectively
AD/BD =AE/CE
such that
and LADE = LACB. Prove that ABC is an isosceles triangle. (please don't spam unnecessarily) It's urgent please help me
Answers
GIVEN: In Δ ABC, D and E are points on AB and AC , DE || BC and AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BE = 5 cm.
In Δ ADE and Δ ABC,
∠ADE =∠ABC (corresponding angles)
[DE || BC, AB is transversal]
∠AED =∠ACB (corresponding angles)
[DE || BC, AC is transversal]
So, Δ ADE ~ Δ ABC (AA similarity)
Therefore, AD/AB = AE/AC = DE/BC
[In similar triangles corresponding sides are proportional]
AD/AB = DE/BC
2.4/(2.4+DB) = 2/5
2.4 × 5 = 2(2.4+ DB)
12 = 4.8 + 2DB
12 - 4.8 = 2DB
7.2 = 2DB
DB = 7.2/2
DB = 3.6 cm
Similarly, AE/AC = DE/BC
3.2/(3.2+EC) = 2/5
3.2 × 5 = 2(3.2+EC)
16 = 6.4 + 2EC
16 - 6.4 = 2EC
9.6 = 2EC
EC = 9.6/2
EC = 4.8 cm
Hence,BD = 3.6 cm and CE = 4.8 cm.
Step-by-step explanation:
Given ∠D=∠E,
DB
AD
=
EC
AE
In triangle ADE, ∠D=∠E
⇒AD=AE ___(1)
Now
DB
AD
=
EC
AE
⇒
AD
DB
=
AE
EC
⇒
AD
DB
+1=
AE
EC
+1
⇒
AD
DB+AD
=
AE
EC+AE
⇒
AD
BA
=
AE
CA
⇒
AE
BA
=
AE
CA
[ from (1)]
⇒BA=CA
⇒△BAC s an isosceles triangle
Hence proved.