Given: In △ABC, AD ⊥ BC and AD² = BD · DC.
To Prove: ∠BAC = 90°
Proof:
Since AD ⊥ BC, we have ∠ADB = ∠ADC = 90°.
In △ADB and △ADC:
Consider △ADB and △CAD.
From the given condition: $AD^2 = BD \cdot DC$
$$\Rightarrow \frac{AD}{DC} = \frac{BD}{AD}$$
Also, ∠ADB = ∠ADC = 90° (since AD ⊥ BC)
Therefore, by SAS similarity criterion:
$$\triangle ADB \sim \triangle CAD$$
(sides about equal angles are proportional, with ∠ADB = ∠CAD... )
Let us use AA criterion instead:
In △ADB and △CAD:
Correct approach:
In △ABD and △ACD, we know AD ⊥ BC (∠ADB = ∠ADC = 90°).
From AD² = BD · DC:
$$\frac{AD}{DC} = \frac{BD}{AD}$$
In △ADB and △CDA:
By SAS similarity: △ABD ~ △CAD
∴ ∠ABD = ∠CAD ... (1)
Similarly, in △ACD and △BAD:
△ACD ~ △BAD (same reasoning)
∴ ∠ACD = ∠BAD ... (2)
Now, ∠BAC = ∠BAD + ∠CAD = ∠ACD + ∠ABD
In △ABC: ∠ABC + ∠ACB + ∠BAC = 180°
∴ ∠BAC + ∠BAC = 180° − ...
Clean final proof:
In △ABD and △CAD:
$$\frac{BD}{AD} = \frac{AD}{CD} \quad \text{(since } AD^2 = BD \cdot DC\text{)}$$
∠BDA = ∠ADC = 90°
By SAS similarity, △ABD ~ △CAD.
∴ ∠BAD = ∠ACD ...(1)
In △ACD and △BAD:
$$\frac{CD}{AD} = \frac{AD}{BD}, \quad \angle ADC = \angle ADB = 90°$$
By SAS similarity, △ACD ~ △BAD.
∴ ∠DAC = ∠ABD ...(2)
Adding (1) and (2):
∠BAD + ∠DAC = ∠ACD + ∠ABD
∴ ∠BAC = ∠ACB + ∠ABC
In △ABC: ∠BAC + ∠ABC + ∠ACB = 180°
∴ ∠BAC + ∠BAC = 180°
∴ 2∠BAC = 180° ⟹ ∠BAC = 90° $\blacksquare$
Source: Chapter 6, Section 6.4 — Criteria for Similarity of Triangles
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