ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively, such that EF || AB. Show that

OR

The perpendicular from A on the side BC of a Δ ABC intersects BC at D, such that DB = 3 CD. Prove that 2 AB² = 2 AC² + BC².

Given: ABCD is a trapezium

where AB || DC

E and F are points on non – parallel sides AD and BC such that EF || AB

To Prove:

Proof:

Given EF || AB and AB || DC

⇒ EF || DC

[Lines which are parallel to same line are parallel to each other]

Construction: Join A to C

Let AC intersect EF at point G

Now, In ΔADC

EG || DC [Because EF || DC]

So, …(i)

[Lines drawn parallel to one side of triangle, intersects the other two sides in distinct points, then it divides the other two sides in same ratio]

Similarly, In Δ CAB

AB || GF [Because EF || AB]

So, …(ii)

[Lines drawn parallel to one side of triangle, intersects the other two sides in distinct points, then it divides the other two sides in same ratio]

From eq. (i) and (ii), we have

Hence Proved

OR

Given: ABC is a triangle

and A is perpendicular on side BC i.e. AD ⊥ BC

Also, DB = 3 CD

To Prove: 2 AB² = 2 AC² + BC²

Proof:

Let BC = x

⇒ CD + DB = x

⇒ CD + 3CD = x [given]

⇒ 4CD = x

…(i)

And DB = 3CD

…(ii)

In Δ ADC, Using Pythagoras theorem

(Perpendicular)² + (Base)² = (Hypotenuse)²

(AD)² + (CD)² = (AC)²

[from (i)]

…(iii)

In Δ ADB, Using Pythagoras theorem

(Perpendicular)² + (Base)² = (Hypotenuse)²

(AD)² + (DB)² = (AB)²

[from (ii)]

…(iv)

From eq. (iii) and (iv), we get

⇒ 2AB² – 2AC² = x²

⇒ 2AB² – 2AC² = BC²

[In starting we let BC = x]

or 2AB² = 2AC² + BC²

Hence Proved