Answer :

  • <B=<E

[tex]\\ \sf\longmapsto ∆AED=∆ABO[/tex]

[tex]\\ \sf\longmapsto \dfrac{AE}{AB}=\dfrac{AD}{AO}[/tex]

[tex]\\ \sf\longmapsto \dfrac{AD}{AE}=\dfrac{AB}{AO}[/tex]

SalmonWorldTopper

Explanation:

Given that

In ∆ ABC, DE is a line drawn on AB and AC

In ∆ ADE and ∆ ABC

∠ AED = ∠ ABC = 70°

∠ AED = ∠ BAC (Common vertex angle )

By AA similarity criteria

∆ ADE and ∆ ABC are similar triangles

⇛ ∆ ADE ~ ∆ ABC

We Know that

If two polygons of same number of sides are said to be similar,

If the corresponding angles are equal.

If the corresponding sides are in the same ratio or in proportion.

So, The corresponding angles are mist be in proportion

⇛ AD / DB = AE/EC

Key Knowledge:-

  • If two polygons of same number of sides are said to be similar,
  • If the corresponding angles are equal.
  • If the corresponding sides are in the same ratio or in proportion.

Note :-

  • If DE || BC then by Basic Proportionality Theorem, AD / DB = AE/EC
  • If a line drawn parallel to one side of a triangle intersecting other two sides at two different points and the line divides the other sides in the same ratio.
${teks-lihat-gambar} SalmonWorldTopper

Other Questions