解答图中几何题---**Extracted Content:**
**Question Stem:**
例 1.如图, 在△ABC中, D是BC边上的一点, AE是△ABD的中线, CD=AB, ∠BDA=∠BAD. 求证: AC=2AE.
(Example 1. As shown in the figure, in △ABC, D is a point on side BC, AE is the median of △ABD, CD=AB, ∠BDA=∠BAD. Prove: AC=2AE.)
**Diagram Description:**
Geometric figure showing triangle ABC. Point D is on side BC. Point E is shown on segment AD, however, segments BE and ED are marked with short horizontal lines, indicating BE = ED, suggesting E is the midpoint of BD. Segment CD and AB are marked with single vertical lines, indicating CD = AB. Angles ∠BDA and ∠BAD are marked with the same arc, indicating ∠BDA = ∠BAD. Angles are labeled numerically and with variables: ∠DAB=x, ∠BDA=x, ∠ABD=180°-2x, ∠ADC=180°-x. Point H is constructed by extending line segment AE such that AE=EH, and segment DH is drawn as a dashed line. Segments AE and EH are marked with double vertical lines, indicating AE = EH.
**Handwritten Notes/Proof Steps:**
在△ABD中
AE是△ABD的中线
延长AE至点H, 使AE=EH, 连接DH
{BE=ED
{AE=EH
∠AEB=∠HED (对顶角相等)
∴△ABE ≅ △HDE (SAS)
∴AB=DH
∠BAD=∠EDH
∵∠BDA=∠BAD (已知)
∴∠BDA=∠EDH (等量代换)
即 ∠ADC = 180°-x
{∠ADH = ∠ADC
在△ADH和△ADC中
{AD = AD (公共边)
{DH = DC (已证)
{∠ADH = ∠ADC
∴△ADH ≅ △ADC (SAS)
∴AH=AC
∵AH=2AE
∴AC=2AE