Prove 2AB² = 2AC² + BC² When DB = 3CD | Class 10 Triangles

Quick answer: If the perpendicular from A meets BC at D with DB = 3CD, then BC = 4CD, so CD = BC/4 and DB = 3BC/4. Pythagoras in triangles ADB and ADC gives AB² − AC² = DB² − CD² = (9/16 − 1/16)BC² = BC²/2. Multiplying by 2 gives 2AB² = 2AC² + BC².

Syllabus note: This was NCERT Exercise 6.5, Question 14. In the rationalised NCERT Class 10 book (from 2023–24), the Pythagoras theorem section and Exercise 6.5 were removed from the Triangles chapter. The proof is still good practice in using Pythagoras, but check your current syllabus before treating it as a board question.

The setup

In triangle ABC, AD is drawn perpendicular to BC and meets it at D. This splits ABC into two right-angled triangles, ADB and ADC, which share the side AD.

            A
           /|\
          / | \
         /  |  \
        /   |   \
       B----D----C
          AD ⊥ BC

The trick in every problem of this kind is the same: write Pythagoras in both right triangles, then eliminate the common side AD.

Theorem 1: AB² + CD² = BD² + AC²

Given: In triangle ABC, AD ⊥ BC.

To prove: AB² + CD² = BD² + AC²

Step 1: In right triangle ADB, by Pythagoras:

$AB^2 = AD^2 + BD^2$

So:

$AD^2 = AB^2 - BD^2 \qquad \text{(1)}$

Step 2: In right triangle ADC, by Pythagoras:

$AC^2 = AD^2 + CD^2$

So:

$AD^2 = AC^2 - CD^2 \qquad \text{(2)}$

Step 3: From (1) and (2), both right sides equal AD²:

$AB^2 - BD^2 = AC^2 - CD^2$

Step 4: Rearrange:

$AB^2 + CD^2 = AC^2 + BD^2$

Hence proved.

Theorem 2: 2AB² = 2AC² + BC² when DB = 3CD

Given: In triangle ABC, AD ⊥ BC and DB = 3CD.

To prove: 2AB² = 2AC² + BC²

Step 1: Write CD and DB in terms of BC.

Since D lies on BC:

$BC = CD + DB = CD + 3CD = 4CD$

So:

$CD = \frac{1}{4}BC, \qquad DB = \frac{3}{4}BC$

Step 2: Pythagoras in both triangles.

In right triangle ADB:

$AB^2 = AD^2 + DB^2 \qquad \text{(1)}$

In right triangle ADC:

$AD^2 = AC^2 - CD^2 \qquad \text{(2)}$

Step 3: Eliminate AD². Put (2) into (1):

$AB^2 = AC^2 + DB^2 - CD^2$

Step 4: Substitute the values of DB and CD.

$AB^2 = AC^2 + \frac{9}{16}BC^2 - \frac{1}{16}BC^2$

$AB^2 = AC^2 + \frac{8}{16}BC^2 = AC^2 + \frac{1}{2}BC^2$

Step 5: Multiply both sides by 2.

$2AB^2 = 2AC^2 + BC^2$

Hence proved.

Check with numbers: Take CD = 1, DB = 3 (so BC = 4) and AD = 2. Then AB² = 4 + 9 = 13 and AC² = 4 + 1 = 5. Left side: 2 × 13 = 26. Right side: 2 × 5 + 16 = 26. ✓

Summary table

Exam tips

• Draw the figure and mark the right angles at D before you start.
• Write BC = CD + DB = 4CD as the very first line. The rest of the proof follows from it.
• Number your equations (1) and (2) and say clearly "substituting (2) in (1)". Examiners give marks for each step.
• Keep the answer in the exact form asked: 2AB² = 2AC² + BC², not AB² = AC² + BC²/2.

Common mistakes

• Squaring only the numerator or only the denominator: (3BC/4)² is 9BC²/16, not 3BC²/16 or 9BC²/4.
• Writing BC = 3CD instead of BC = 4CD (forgetting to add CD itself).
• Using Pythagoras on triangle ABC, which is not right-angled.
• Mixing up which segment is three times the other. Here DB is the longer part, so AB is the longer side.

Frequently Asked Questions

How do you prove 2AB² = 2AC² + BC²?

Use DB = 3CD to get CD = BC/4 and DB = 3BC/4. Apply Pythagoras to triangles ADB and ADC, eliminate AD² to get AB² − AC² = DB² − CD² = BC²/2, then multiply by 2.

Why is BC equal to 4CD?

Because D lies on BC, so BC = CD + DB. Since DB = 3CD, BC = CD + 3CD = 4CD.

How do you prove AB² + CD² = BD² + AC²?

Write AD² = AB² − BD² from triangle ADB and AD² = AC² − CD² from triangle ADC, equate the two and rearrange.

Is this question still in the Class 10 NCERT syllabus?

It was Exercise 6.5, Question 14 in the older NCERT book. The rationalised book removed the Pythagoras theorem section from the Triangles chapter, so check your school's current syllabus.

Which theorem is used in this proof?

Only the Pythagoras theorem, applied to the two right-angled triangles formed by the perpendicular AD.